Equal volumes of 0.2M HCI and 0.4M KOH are mixed. The concentration of ions in the resulting solution are:

Options

A. `[K^(+) = 0.4 M , [C1^(-) ] = 0.2 M , [H^(+)] = 0.2 M`
B. `[K^(+) ] = 0.2 M , [C1^(-)]=0.1M, [OH^(-)] = 0.1 M`
C. `[K^(+)]=0.1 M, [C1^(-)]=0.1 M, [OH^(-)]=0.1M`
D. `[K^(+)]=0.2M, [C1^(-)]=0.1M, [OH^(-)]=0.2M`

Answer

To solve the problem of finding the concentration of ions in the resulting solution when equal volumes of 0.2M HCl and 0.4M KOH are mixed, we will follow these steps:

### Step 1: Calculate the moles of HCl and KOH
Since we are mixing equal volumes of both solutions, let's assume we take a volume \( V \) liters for each.

- Moles of HCl:
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 0.2 \, \text{M} \times V = 0.2V \]

- Moles of KOH:
\[ \text{Moles of KOH} = \text{Molarity} \times \text{Volume} = 0.4 \, \text{M} \times V = 0.4V \]

### Step 2: Determine the dissociation of HCl and KOH
HCl dissociates completely in water:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
From 0.2V moles of HCl, we get:
- Moles of \( \text{H}^+ \) = 0.2V
- Moles of \( \text{Cl}^- \) = 0.2V

KOH also dissociates completely:
\[ \text{KOH} \rightarrow \text{K}^+ + \text{OH}^- \]
From 0.4V moles of KOH, we get:
- Moles of \( \text{K}^+ \) = 0.4V
- Moles of \( \text{OH}^- \) = 0.4V

### Step 3: Calculate the total volume of the solution
The total volume of the mixed solution is:
\[ \text{Total Volume} = V + V = 2V \]

### Step 4: Calculate the concentrations of each ion
Now we can calculate the molarity (concentration) of each ion in the final solution:

- Molarity of \( \text{H}^+ \):
\[ \text{Molarity of } \text{H}^+ = \frac{\text{Moles of } \text{H}^+}{\text{Total Volume}} = \frac{0.2V}{2V} = 0.1 \, \text{M} \]

- Molarity of \( \text{Cl}^- \):
\[ \text{Molarity of } \text{Cl}^- = \frac{\text{Moles of } \text{Cl}^-}{\text{Total Volume}} = \frac{0.2V}{2V} = 0.1 \, \text{M} \]

- Molarity of \( \text{K}^+ \):
\[ \text{Molarity of } \text{K}^+ = \frac{\text{Moles of } \text{K}^+}{\text{Total Volume}} = \frac{0.4V}{2V} = 0.2 \, \text{M} \]

- Molarity of \( \text{OH}^- \):
\[ \text{Molarity of } \text{OH}^- = \frac{\text{Moles of } \text{OH}^-}{\text{Total Volume}} = \frac{0.4V}{2V} = 0.2 \, \text{M} \]

### Step 5: Summary of concentrations
The final concentrations of the ions in the resulting solution are:
- \( \text{H}^+ \) = 0.1 M
- \( \text{Cl}^- \) = 0.1 M
- \( \text{K}^+ \) = 0.2 M
- \( \text{OH}^- \) = 0.2 M